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Prove that the sequence $x_n = 1 +\frac{ sin (n+ \pi) }{n} $ is a cauchy sequence using the definition:

$$\forall \epsilon>0 \exists N\in\mathbb{N}: n,m\ge N\implies |x_n-x_m|<\epsilon.$$

I have tried to prove: $ | \frac{ n-sin(n)}{n} - \frac{ m -sin(m)}{m} | \leq \epsilon$. The triangle inequality did not work for me and I don't know how to prove it with the provided definition.

I could argue that the sequence converges to 1 and is therefore cauchy. However, I need to prove this with the definition.

Please help me.

smalllearner
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2 Answers2

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Note that\begin{align}|x_m-x_n|&=\left|1+\frac{\sin(m+\pi)}m-\left(1+\frac{\sin(n+\pi)}n\right)\right|\\&=\left|\frac{\sin(m+\pi)}m-\frac{\sin(n+\pi)}n\right|\\&\leqslant\frac1m+\frac1n.\end{align}So, given $\varepsilon>0$, take $N\in\Bbb N$ such that $\frac1N<\frac\varepsilon2$ and then$$m,n\geqslant N\implies|x_m-x_n|\leqslant\frac1m+\frac1n\leqslant\frac2N<\varepsilon.$$

José Carlos Santos
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  • Hello thank you very much. Could you please tell me why it's $\leq 1/m + 1/n$? I'm learning on my own and I don't know how we got there. Thanks again. – smalllearner May 08 '20 at 07:23
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    \begin{align}|x_m-x_n|&=\left|\frac{\sin(m+\pi)}m-\sin{(n+\pi)}n\right|\&\leqslant\left|\frac{\sin(m+\pi)}m\right|+\left|\frac{\sin(n+\pi)}n\right|\text{ (triangle inequality)}\&\leqslant\frac1m+\frac1n\end{align} – José Carlos Santos May 08 '20 at 07:49
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$|\frac{\sin (π+m)}{m}-\frac{\sin (π+n)}{n}|\le$

$|\frac{\sin (π+m)}{m}| +|\frac{\sin (π+n)}{n}|\le$

$1/m+1/n;$

Let $\epsilon >0$ be given.

Choose $n_0 > 2/\epsilon$ (Archimedean principle).

For $m\ge n \ge n_0:$

$|\frac{\sin (π+m)}{m}-\frac{\sin (π+n)}{n}|\le 1/m+1/n \le 2/n \le 2/n_0 <\epsilon.$

Peter Szilas
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